- Blowdown rate
- 909 lb/h
- Blowdown, % of feedwater
- 8.3 %
- Cycles of concentration
- 12.0
- Heat in blowdown water
- 281,545 Btu/h
- Annual heat cost (before recovery, at 100 % efficiency)
- $13,514
The short answer
Blowdown rate = steam rate × feedwater TDS ÷ (maximum boiler TDS − feedwater TDS). For example, a boiler making 10,000 lb/h of steam with 250 ppm feedwater and a 3,000 ppm boiler limit needs about 909 lb/h of blowdown, or 8.3 % of feedwater, at 12 cycles of concentration.
Formula
Blowdown (lb/h) = S × F ÷ (B − F)Cycles of concentration = B ÷ FBlowdown as % of feedwater = F ÷ B × 100Heat lost (Btu/h) ≈ blowdown × (h_boiler water − h_make-up)
How to use it
- Enter the steam rate.
- Enter feedwater TDS (or conductivity) and the boiler-water TDS limit your water-treatment program sets, in the same units.
- Add the boiler pressure and make-up temperature to see the heat in the blowdown water.
Assumptions
- S = steam rate, F = feedwater TDS, B = maximum boiler-water TDS. Conductivity can be used instead of TDS if both are in the same units.
- Heat loss uses saturated-water enthalpy at boiler pressure and make-up water enthalpy; a flash-steam recovery or blowdown heat exchanger recovers part of it.
Frequently asked questions
What is a typical blowdown rate?
Between about 1 % and 8 % of feedwater flow for most packaged steam boilers. Higher rates usually mean poor make-up water quality or little condensate return.
What happens if blowdown is too low?
Dissolved solids build up, which causes foaming, carryover into the steam, and scale on heat-transfer surfaces.
Continuous or bottom blowdown?
Continuous (surface) blowdown controls TDS. Bottom blowdown removes sludge from the lowest point and is done in short, regular bursts as the boiler maker and water-treatment specialist specify.